By M. Fremond, S. Miyazaki (auth.)
This e-book comprises chapters. the 1st bankruptcy bargains with the thermomechanical macroscopic thought describing the transformation and deformation habit of form reminiscence alloys. the second one bankruptcy offers with the huge and basic evaluation of the experimental works which come with crystallography, variations and mechanical features in Ti-Ni, Cu-base and ferrous form reminiscence alloys.
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Extra info for Shape Memory Alloys
A Thermomechanical Macroscopic Theory 57 Figure 28 Constitutive law of a non-dissipative educated shape memory alloy at medium temperature. 5. The two shape memory effect. Let us consider an unloaded (cr = 0) educated shape memory alloy at high temperature, T+. We have E = 0. Let us cool it, the only unloaded (cr = 0) equilibrium state at low remperature, T-, is (cr =0, E =E2}. If the temperature is again increased the alloys goes back to the state (cr =0, E =0) and so on (figure 29). The alloys remembers two shapes: the state Er+(E = 0, cr = 0, ~3 = 1, T+) and the state Er-(E = Ez, cr = 0, p1 = 1, T-) !
Evolution of a shape memory allov. The dissipative terms of the constitutive law (52) or of the relation (52bis) involve derivatives with respect to the time. Thus they are differential equations. The natural problem to look at is the evolution of a material submitted to time dependant external actions. In this paragraph we study unidimensional experiments. We choose to apply the deformation E(t) because it is more easy to inverstigate the structure of the equations and exhibit the hysteretical properties of shape memory alloys.
There is only austenite, P3 = 1, and from (48), cr = K€ (figure 8). A Thermomechanical Macroscopic Theory 33 Figure 8 The constitutive low at medium temperature. ForE= E1, we have Yri(Er) = 0. The vector -Yr(Er) is normal to the triangle Cr on the side OA (figure 7) and ~ 1+~3 = l: we have a mixture of austenite and martensite number one. The stress cr can take any value of the segment [KE 1, KE 1-t 11 (T)] (figure 8). ForE> EJ, we have -Yri(Er) > 0 and -Yr2(Er) < 0. The vector -Yr(Er) is normal to the triangle Cr at the vertex A (figure 7) and~~ = l, there is only the martensite number one and the stress is cr = KE-'tii(T).